bashshellargument-passingautomator

Remove last argument from argument list of shell script (bash)


This question concerns a bash script that is run in automator osx. I am using automator actions to get and filter a bunch of file references from the finder. Then I append to that list the name of the parent folder, also via an automator action. Automator then feeds these arguments to an action called "run shell script". I am not sure exactly how automator invokes the script but the argument list looks like this when echoed with: echo "$@"

/Volumes/G-Raid/Online/WAV_TEST/Testbok 50/01/01000 43-001.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/02/02000 43-002.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/03/03000 43-003.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50

In this case path to 3 files and a folder.

In the shell script I launch an application called ripcheckc* with the args passed from automator minus the last argument(the folder) in the list.

I use this to remove the last argument:

_args=( "$@" )
unset _args[${#_args[@]}-1]

And this is echo $_args:

/Volumes/G-Raid/Online/WAV_TEST/Testbok 50/01/01000 43-001.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/02/02000 43-002.wav /Volumes/G-Raid/Online/WAV_TEST/Testbok 50/03/03000 43-003.wav

Same as before but without the folder.

Now, if I run ripcheckc with "$@" as argument it works (but fails later on because of that last path in the argument list) If I use ${_args[@]} the application will just abort silently. When I echo $@ and _args the output looks identical except for the last argument.

My question is - what is the difference between $@ and $_args that make the first valid input and the second not?

*The application is ripcheckc

I hope my question makes sense.

EDIT: Solved.


Solution

  • Assuming that you already have an array, you can say:

    unset "array[${#array[@]}-1]"
    

    For example, if your script contains:

    array=( "$@" )
    unset "array[${#array[@]}-1]"    # Removes last element -- also see: help unset
    for i in "${array[@]}"; do
      echo "$i"
    done
    

    invoking it with: bash scriptname foo bar baz produces:

    foo
    bar